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How do I solve this equation. Find a and b for a given prime p. What properties must p have for a solution to this equation to exist. a^2 + b^2 = 0 (mod p) and

Message 1 of 19
, Feb 2, 2004

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How do I solve this equation. Find a and b for a given prime p.
What properties must p have for a solution to this equation to exist.

a^2 + b^2 = 0 (mod p)

and what if b= a+1

Let me know!

Thanks,
Harsh Aggarwal

Carl Devore

... Not much interesting about this. Since a^2 = -b^2 (mod p), we have that solutions exist iff -1 is a quadratic residue mod p. If it is, let i denote a

Message 2 of 19
, Feb 2, 2004

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On Tue, 3 Feb 2004, eharsh82 wrote:

> How do I solve this equation. Find a and b for a given prime p.
> What properties must p have for a solution to this equation to exist.
> a^2 + b^2 = 0 (mod p)

Not much interesting about this. Since a^2 = -b^2 (mod p), we have that
solutions exist iff -1 is a quadratic residue mod p. If it is, let i
denote a square root of -1 (which can be found by the algorithm we
discussed here a few weeks ago). Then a = +/- i*b for any b are
solutions.

> and what if b= a+1

Then it becomes the quadratic equation 2*a^2+2*a+1 = 0, which can be
solved by the algorithm discussed a few weeks ago.

mikeoakes2@aol.com

... This problem is not very well posed. Solutions to that equation exist for /any/ prime p, as all it says is that p divides the lhs! I think you want the

> How do I solve this equation. Find a and b for a given prime p.
> What properties must p have for a solution to this equation to exist.
>
> a^2 + b^2 = 0 (mod p)
>

This problem is not very well posed.
Solutions to that equation exist for /any/ prime p, as all it says is that p
divides the lhs!

I think you want the restrictions that a, b are in [0,p-1], no?

In that case, either
(a) p = 2, or
(b) p = 1 mod 4.

Case (a) gives as the unique solution a=b=1.

Case (b) has many solutions.
Choose a = any integer in [1,p-1];
then b = a*e mod p,
where e is one of the 2 solutions to the congruence
e^2 = -1 mod p.

I think there were recent posts to this group about how to take square roots
mod p.

-Mike Oakes

[Non-text portions of this message have been removed]

eharsh82

I would like to call primes of the form a^2+b^2=prime, pythagorean primes. Can anyone proove that there are infinite pythagorean primes? How about the

Message 4 of 19
, Feb 3, 2004

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I would like to call primes of the form a^2+b^2=prime, pythagorean
primes.

Can anyone proove that there are infinite pythagorean primes?
How about the distribution of a prime p where a^2+b^2=p^2 ?
What if b=a+1, then do infinite pytahgorean primes exist?

I think something similar to how Euler prooved there are infinite
primes may work here but I am not sure.

Let me know!

Thanks,
Harsh Aggarwal

--- In primenumbers@yahoogroups.com, "eharsh82" <harsh@u...> wrote:
> How do I solve this equation. Find a and b for a given prime p.
> What properties must p have for a solution to this equation to
exist.
>
> a^2 + b^2 = 0 (mod p)
>
> and what if b= a+1
>
> Let me know!
>
> Thanks,
> Harsh Aggarwal

Andrew Swallow

... Or you could just call them Gaussian primes, which is more or less what they are. You should be able to find necessary information in any introductory

> I would like to call primes of the form a^2+b^2=prime, pythagorean
> primes.
>
> Can anyone proove that there are infinite pythagorean primes?
> How about the distribution of a prime p where a^2+b^2=p^2 ?
> What if b=a+1, then do infinite pytahgorean primes exist?
>
> I think something similar to how Euler prooved there are infinite
> primes may work here but I am not sure.

Or you could just call them Gaussian primes, which is more or less
what they are. You should be able to find necessary information in any
introductory number theory book. Or on whatever websites there are,
probably.

As for b=a+1, well that changes it to just a single dimensional
problem, and is probably more difficult to answer.

eharsh82

If both a and b are nonzero then z=a+bi, is a Gaussian prime iff a^2+b^2 is an ordinary prime. So there is actually no name for primes of the form a^2+b^2

Message 6 of 19
, Feb 3, 2004

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If both a and b are nonzero then z=a+bi, is a Gaussian prime iff
a^2+b^2 is an ordinary prime.

> > primes.
> >
> > Can anyone proove that there are infinite pythagorean primes?
> > How about the distribution of a prime p where a^2+b^2=p^2 ?
> > What if b=a+1, then do infinite pytahgorean primes exist?
> >
> > I think something similar to how Euler prooved there are infinite
> > primes may work here but I am not sure.
>
> Or you could just call them Gaussian primes, which is more or less
> what they are. You should be able to find necessary information in

any

> introductory number theory book. Or on whatever websites there are,
> probably.
>
> As for b=a+1, well that changes it to just a single dimensional
> problem, and is probably more difficult to answer.

elevensmooth

... Every prime of the form 4n+1 is the sum of two squares. Euler first communicated the following elegant proof of this fact to Goldbach in 1749, two years

> I would like to call primes of the form a^2+b^2=prime, pythagorean
> primes.
>
> Can anyone proove that there are infinite pythagorean primes?

"Every prime of the form 4n+1 is the sum of two squares. Euler first
communicated the following elegant proof of this fact to Goldbach in
1749, two years after his original proof which was rathar vague on
this point ..."

> > primes.
> >
> > Can anyone proove that there are infinite pythagorean primes?
>
> "Every prime of the form 4n+1 is the sum of two squares. Euler

first

> communicated the following elegant proof of this fact to Goldbach in
> 1749, two years after his original proof which was rathar vague on
> this point ..."
>
> Fermat's Last Theorem by Edwards, Springer Verlag, 1977.
> --
> ElevenSmooth: Distributed Factoring of 2^3326400-1
> http://ElevenSmooth.com

Carl Devore

... Then it becomes a univariate quadratic polynomial with interger coefficients. No univariate polynomial of degree greater than has ever been proved to

Message 9 of 19
, Feb 3, 2004

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On Wed, 4 Feb 2004, eharsh82 wrote:

> So what about when b=a+1

Then it becomes a univariate quadratic polynomial with interger
coefficients. No univariate polynomial of degree greater than has ever
been proved to produce an infinite number of primes when integers are
substituted for the variable.

Carl Devore

... Should say ...degree greater than one has ever...

Message 10 of 19
, Feb 3, 2004

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On Tue, 3 Feb 2004, Carl Devore wrote:

> Then it becomes a univariate quadratic polynomial with interger
> coefficients. No univariate polynomial of degree greater than has ever
> been proved to produce an infinite number of primes when integers are
> substituted for the variable.

Should say "...degree greater than one has ever..."

eharsh82

I am not sure if this series has finite number of primes or not. I think it has infinite primes. I have found several primes in the 10000 digit category. I

Message 11 of 19
, Feb 3, 2004

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I am not sure if this series has finite number of primes or not.
I think it has infinite primes.

I have found several primes in the 10000 digit category.
I used PFGW on (10^10000+$a)^2+ (10^10000+$a+1)^2

I'm also working on the series:
(2^$a)^2+(2^$a+1)^2
and
(2^$a)^2+(2^$a-1)^2

> > substituted for the variable.
>
> Should say "...degree greater than one has ever..."

Andy Swallow

... But that is a question you will never be able to answer, one way or another, if all you re doing is search for primes of this type using computer methods.

Message 12 of 19
, Feb 4, 2004

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On Wed, Feb 04, 2004 at 05:31:24AM -0000, eharsh82 wrote:

> I am not sure if this series has finite number of primes or not.
> I think it has infinite primes.

But that is a question you will never be able to answer, one way or
another, if all you're doing is search for primes of this type using
computer methods. So wouldn't it be more interesting to study the
abstract theory? Your original question was about Gaussian primes, or
primes congruent to 1 mod 4. That's all interesting and fairly basic
stuff. I would have thought that more informative answers would be found
in there.

Apologies if I'm talking rubbish. It just seems strange that on the one
hand you're interested in whether certain sets contain infinitely many
primes, yet on the other hand you're studying the sets using methods
guaranteed to not be able to answer the question, :-)

Anyway, that's my morning rant out of the way...

Andy

eharsh82

Here is my proof for infiniteness of these primes if b=a+1 then we get 2*a^2+2*a+1 =p solving this a is an integer if there is a prime p such that 2*p-1=m^2 or

Message 13 of 19
, Feb 4, 2004

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Here is my proof for infiniteness of these primes

if b=a+1
then we get 2*a^2+2*a+1 =p
solving this
a is an integer if there is a prime p such that 2*p-1=m^2
or m^2+1/2 is a prime

The distribution of such primes would follow the distribution of
primes with the formula n^2+1

it is conjectured that such primes are infinite.

----
Taken from primepages.com

Are there infinitely many primes of the form n2+1?
There are infinitely many of the forms n2+m2 and n2+m2+1. A more
general form of this conjecture is if a, b, c are relatively prime, a
is positive, a+b and c are not both even,and b2-4ac is not a perfect
square, then there are infinitely many primes an2+bn+c [HW79, p19].

---

What do you all think?

Also the series I talked about, what do you think about's it
distribution.

> On Wed, Feb 04, 2004 at 05:31:24AM -0000, eharsh82 wrote:
> > I am not sure if this series has finite number of primes or not.
> > I think it has infinite primes.
>
> But that is a question you will never be able to answer, one way or
> another, if all you're doing is search for primes of this type using
> computer methods. So wouldn't it be more interesting to study the
> abstract theory? Your original question was about Gaussian primes,

or

> primes congruent to 1 mod 4. That's all interesting and fairly basic
> stuff. I would have thought that more informative answers would be

found

> in there.
>
> Apologies if I'm talking rubbish. It just seems strange that on the

one

> hand you're interested in whether certain sets contain infinitely

many

> primes, yet on the other hand you're studying the sets using methods
> guaranteed to not be able to answer the question, :-)
>
> Anyway, that's my morning rant out of the way...
>
> Andy

pop_stack

Hi, I m a newby so please be gentle. OK There is no integer solution for X^2 + Y^2 = Z^2 when X and Y are prime. (In other words, no two primes, squared and

Message 14 of 19
, Feb 9, 2004

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Hi, I'm a newby so please be gentle.
OK

There is no integer solution for
X^2 + Y^2 = Z^2 when X and Y are prime.

(In other words, no two primes, squared and sumed can equal a perfect
square).

This came out of my Pythagorean triplets program
that seems to show that either X AND/OR Z are always prime
and Y is never prime.

I am not sophisticated enough to know whether the above is trivial.

But, I'm excited to find a group for prime numbers.

BTW, I have Visual Basic and or Excel demonstrations of
the statements above. No proofs, of course.

... This holds if x and y are both odd (not just prime) because x^2+y^2 must be congruent to 2 mod 4 and z^2 must be congruent to 0 mod 4. It holds when x=2

Message 15 of 19
, Feb 9, 2004

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At 11:30 PM 2/9/2004, pop_stack wrote:

>Hi, I'm a newby so please be gentle.
>OK
>
>There is no integer solution for
>X^2 + Y^2 = Z^2 when X and Y are prime.

This holds if x and y are both odd (not just prime) because x^2+y^2 must be
congruent to 2 mod 4 and z^2 must be congruent to 0 mod 4. It holds when
x=2 (more generally when x == 2 mod 4) and y is odd for similar reasons.

eharsh82

I realized that the series (2^n+1)^2 + (2^n)^2 and (2^n-1)^2 + (2^n) ^2 are nothing but Aurifeuillian Factors. They have some special properties that I have

Message 16 of 19
, Feb 25, 2004

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I realized that the series (2^n+1)^2 + (2^n)^2 and (2^n-1)^2 + (2^n)
^2 are nothing but Aurifeuillian Factors.

They have some special properties that I have discovered.

1) Basically these numbers are 2^p+2^((p+1)/2)+1 and 2^p-2^((p+1)/2)+1

3) Either L or M is divisible by 5. When p%8=1 or 5 M is divisible by
5 and when p%8=3 or 7 L is divisible by 5. So only one candidate
remains for each prime.

4) So the factors of these numbers are of the form 4*p*k+1.

5) I think their distribution is really similar to mersenne primes,
as most of their properties. I have searched these numbers up to
p=35000 and am continuing to search higher. I have found them to
produce an equal number of primes as mersenne numbers. I think these
primes are the Gaussian equivalents of mersenne primes.

6) I think a top 20 list of these numbers can be started on the
primepages.org web page since these numbers are well known and have
been discussed in a lot of papers.

7) I am not sure if DWT can be used productively, with this series.
But if anyone knows how it can be used productively, please let me
know.

In order to speed up the search to higher n's, I am looking for a
sieve/ Trial factorer. Could someone with the required skill please
write me a program to sieve? I did try myself to write one but it is
not very fast. I am currently using that and sieving all numbers up
to 25G before moving to PRPing. (takes about 30 sec to take a
candidate to 25 G)

--- In primenumbers@yahoogroups.com, "eharsh82" <harsh@u...> wrote:
> Here is my proof for infiniteness of these primes
>
> if b=a+1
> then we get 2*a^2+2*a+1 =p
> solving this
> a is an integer if there is a prime p such that 2*p-1=m^2
> or m^2+1/2 is a prime
>
> The distribution of such primes would follow the distribution of
> primes with the formula n^2+1
>
> it is conjectured that such primes are infinite.
>
> ----
> Taken from primepages.com
>
> Are there infinitely many primes of the form n2+1?
> There are infinitely many of the forms n2+m2 and n2+m2+1. A more
> general form of this conjecture is if a, b, c are relatively prime,
a
> is positive, a+b and c are not both even,and b2-4ac is not a
perfect
> square, then there are infinitely many primes an2+bn+c [HW79, p19].
>
> ---
>
>
> What do you all think?
>
> Also the series I talked about, what do you think about's it
> distribution.
>
> a=2^n or 2^n-1
> b=a+1 = 2^n+1 or 2^n
>
> the series reduce to ((2^n-1)^2+1)/2 and ((2^n+1)^2+1)/2
>
> What about the distribution of primes in such a series?
>
> let me know!
>
>
> Harsh Aggarwal
>
>
>
> --- In primenumbers@yahoogroups.com, Andy Swallow
<umistphd2003@y...>
> wrote:
> > On Wed, Feb 04, 2004 at 05:31:24AM -0000, eharsh82 wrote:
> > > I am not sure if this series has finite number of primes or not.
> > > I think it has infinite primes.
> >
> > But that is a question you will never be able to answer, one way
or
> > another, if all you're doing is search for primes of this type
using
> > computer methods. So wouldn't it be more interesting to study the
> > abstract theory? Your original question was about Gaussian
primes,
> or
> > primes congruent to 1 mod 4. That's all interesting and fairly
basic
> > stuff. I would have thought that more informative answers would
be
> found
> > in there.
> >
> > Apologies if I'm talking rubbish. It just seems strange that on
the
> one
> > hand you're interested in whether certain sets contain infinitely
> many
> > primes, yet on the other hand you're studying the sets using
methods
> > guaranteed to not be able to answer the question, :-)
> >
> > Anyway, that's my morning rant out of the way...
> >
> > Andy

mikeoakes2@aol.com

... You are confirming known results. I introduced these Gaussian Mersennes in a post to the Mersenne mailing list in 2000:

> I realized that the series (2^n+1)^2 + (2^n)^2 and (2^n-1)^2 + (2^n)
> ^2 are nothing but Aurifeuillian Factors.
>
> They have some special properties that I have discovered.
>
> 1) Basically these numbers are 2^p+2^((p+1)/2)+1 and 2^p-2^((p+1)/2)+1
>
> 2^(2p)+1=(2^p+2^((p+1)/2)+1)*(2^p-2^((p+1)/2)+1)
> 2^(2p)+1= M2p * L2p (Notation used in literature)
>
> 2) p must be prime so that either L or M can be a base 2-PRP.
>
> 3) Either L or M is divisible by 5. When p%8=1 or 5 M is divisible by
> 5 and when p%8=3 or 7 L is divisible by 5. So only one candidate
> remains for each prime.
>
> 4) So the factors of these numbers are of the form 4*p*k+1.
>
> 5) I think their distribution is really similar to mersenne primes,
> as most of their properties. I have searched these numbers up to
> p=35000 and am continuing to search higher. I have found them to
> produce an equal number of primes as mersenne numbers. I think these
> primes are the Gaussian equivalents of mersenne primes.
>
> 6) I think a top 20 list of these numbers can be started on the
> primepages.org web page since these numbers are well known and have
> been discussed in a lot of papers.
>
> 7) I am not sure if DWT can be used productively, with this series.
> But if anyone knows how it can be used productively, please let me
> know.
>
> In order to speed up the search to higher n's, I am looking for a
> sieve/ Trial factorer. Could someone with the required skill please
> write me a program to sieve? I did try myself to write one but it is
> not very fast. I am currently using that and sieving all numbers up
> to 25G before moving to PRPing. (takes about 30 sec to take a
> candidate to 25 G)
>
> Let me know, if any one can help.
>

I looked at your link and it is quite interesting. I have a couple of
comments though. You mention that G0(n) = n*(1+i)^n + 1 and is
related to Cullens but later state that G0(n) = n*2^(n/2) + 1 and then
go on to show primes of that form. Am I missing something? n*(1+i)^n
+ 1 =/= n*2^(n/2) + 1. The same could be said of G2(n) and Woodalls.

You also have Ne(n) as n^2*2^n + 1, which pre-dates the Hyper-Cullen
search of Steven Harvey. He has noted your finds as he searches up to
200000.

--Mark

mikeoakes2@aol.com

In a message dated 26/02/04 14:20:52 GMT Standard Time, mgrogue@wi.rr.com ... It is if n = 0 mod 8, which was (one of) the values I was talking about.

> You mention that G0(n) = n*(1+i)^n + 1 and is
> related to Cullens but later state that G0(n) = n*2^(n/2) + 1 and then
> go on to show primes of that form. Am I missing something? n*(1+i)^n
> + 1 =/= n*2^(n/2) + 1.
>

It is if n = 0 mod 8, which was (one of) the values I was talking about.
Remember: (1+i)^2 = 2*i.

-Mike Oakes

[Non-text portions of this message have been removed]

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